LeetCode #1010. Pairs of Songs With Total Durations Divisible by 60

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本文探讨了一种算法，用于找出歌曲列表中所有时长总和能被60整除的歌曲对。通过使用哈希表，算法有效地计算出满足条件的歌曲对数量，展示了在大规模数据集上的高效性。

题目描述：

In a list of songs, the i-th song has a duration of time[i] seconds.

Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i < j with (time[i] + time[j]) % 60 == 0.

Example 1:

Input: [30,20,150,100,40]
Output: 3
Explanation: Three pairs have a total duration divisible by 60:
(time[0] = 30, time[2] = 150): total duration 180
(time[1] = 20, time[3] = 100): total duration 120
(time[1] = 20, time[4] = 40): total duration 60

Example 2:

Input: [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.

Note:

1 <= time.length <= 60000
1 <= time[i] <= 500

class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& time) {
        unordered_map<int,int> hash;
        int result=0;
        for(int t:time)
        {
            if(t%60==0) 
            {
                hash[60]++;
                continue;
            }
            t=t%60;
            if(hash.count(60-t)) result+=hash[60-t];
            hash[t]++;
        }
        
        int n=hash[60];
        if(n>=2) result+=n*(n-1)/2;
        return result;
    }
};